What are the vertex, axis of symmetry, maximum or minimum value, domain, and range of the function, and x and y intercepts for # y=x^2-10x+2#?

1 Answer
Apr 29, 2015
  • #y=x^2-10x+2# is the equation of a parabola which will open upwards(because of the positive coefficient of #x^2#)
    So it will have a Minimum

  • The Slope of this parabola is
    #(dy)/(dx) = 2x-10#
    and this slope is equal to zero at the vertex
    #2x - 10 = 0#
    #-> 2x = 10 -> x = 5#

  • The X coordinate of the vertex will be #5#

#y=5^2-10(5)+2 = 25-50+2 =-23#
The vertex is at #color(blue)((5,-23)#

and has a Minimum Value #color(blue)(-23# at this point.

  • The axis of symmetry is #color(blue)(x=5#

  • The domain will be #color(blue)(inRR#(all real numbers)

  • The range of this equation is #color(blue)({y in RR : y>=-23}#

  • To get the x intercepts, we substitute y = 0
    #x^2-10x+2 = 0#
    We get two x intercepts as #color(blue)((5+sqrt23) and (5-sqrt23)#

  • To get the Y intercepts, we substitute x = 0
    # y = 0^2 -10*0 + 2 = 2#
    We get the Y intercept as #color(blue)(2#

  • This is how the Graph will look:
    graph{x^2-10x+2 [-52.03, 52.03, -26, 26]}